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Linear depedency

  1. May 29, 2009 #1
    Prove whethere is linearly independent or not the following:

    {[tex]e^x,e^{2x}[/tex]}
     
  2. jcsd
  3. May 29, 2009 #2

    jbunniii

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    If not, then there must exist two scalars, a and b (they can't both be zero), such that

    [tex]a e^x + b e^{2x} = 0[/tex]

    for all x. Can you find such scalars?
     
  4. May 29, 2009 #3

    HallsofIvy

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    Since that is to be true for all x, you can get two simple equations to solve for a and b by choosing two values of x.
     
  5. May 29, 2009 #4
    I think it's possible:
    [tex]a e^x + b e^{2x} = 0[/tex] <==> [tex]a e^x = - b e^{2x}[/tex] <==> [tex]ln (a e^x) = - ln (b e^{2x})[/tex] <==> ax = -2bx

    So if we carefully choose a, we can definitely show your given set is linearly dependent.
     
  6. May 29, 2009 #5

    jbunniii

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    Except

    [tex]\ln(a e^x) = \ln(a) + x[/tex]
    [tex]\ln(-b e^{2x}) = \ln(-b) + 2x[/tex]

    so you need

    [tex]\ln(a) - \ln(-b) = x[/tex] for all x!
     
  7. May 29, 2009 #6
    Bummer, so sorry.
     
  8. May 29, 2009 #7
    Put x=0,then a+b=0 ====> a=-b

    Hence the above are linearly dependent

    CORRECT???

    THANKS
     
  9. May 30, 2009 #8

    matt grime

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    No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.
     
  10. May 30, 2009 #9

    Mat grime ,please ,write down for me the definition ,when a set of n vectors are linearly independent
     
  11. May 30, 2009 #10

    matt grime

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    Apologies, I missed out an 'in'. The functions e^x and e^2x are clearly linearly independent (over R), which is where I misunderstood what you're saying. (If you don't see it then just rearrange ae^x +be^2x=0 to see that you're claim of linear dependence implies that e^x = -a/b for all x.)
     
    Last edited: May 30, 2009
  12. May 30, 2009 #11

    HallsofIvy

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    No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.

    Another way to prove it is this: if [itex]ae^x+ be^{2x}= 0[/itex] for all x, then it is a constant and its derivative must be 0 for all x: [itex]ae^x+ 2be^{2x}= 0[/itex]. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.
     
  13. May 30, 2009 #12
    Is not the definition for linear indepedence the following:

    for all a,b and x ,real Nos and [tex]ae^x + be^{2x}=0\Longrightarrow a=0=b[/tex]
     
  14. May 30, 2009 #13

    HallsofIvy

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    Yes, that is why you can't "prove [itex]e^x[/itex] and [itex]e^{2x}[/itex] are dependent": they are independent.
     
  15. May 30, 2009 #14

    matt grime

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    Remember this is equal to the zero *function* in the vector space of real valued functions. The zero function is the one that is zero for all its inputs.
     
  16. May 30, 2009 #15


    Then the negation of the above definition should imply for linear dependency

    But the negation of the above definition is:

    There exist a.b, and ,x such that [tex]ae^x + be^{2x} = 0[/tex] and[tex]a\neq 0[/tex] and [tex]b\neq 0[/tex].

    HENCE if we put a=2 ,b=-2 ,x=0 ,we satisfy the linear dependency definition
     
  17. May 31, 2009 #16

    matt grime

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    That isn't right. You have your quantifiers all kludged up into one it should have read

    (for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

    the negation of which is

    (there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

    The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.
     
    Last edited: May 31, 2009
  18. May 31, 2009 #17
    Can you put that into a quantifier form??
     
  19. May 31, 2009 #18

    matt grime

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    In what way is that not in a useful quantifier form (for those who like things like that)? Really, we shouldn't even bother with the quantifier "for all a,b", and putting things in the full on formal abstract quantifier notation just makes things far more opaque than they need to be.

    You're making a very easy question seem very hard: can you find real numbers a and b not both equal to 0 so that

    ae^x + be^2x

    is the zero function? No - it has been proven several times in this thread.
     
  20. May 31, 2009 #19

    HallsofIvy

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    The point is that you are working in a vector space of functions. When we say "af(x)+ bg(x)= 0" we mean it is equal to the 0 function- 0 for all x.
     
  21. May 31, 2009 #20
    The negation of :

    (for all a,b)(ae^x + be^2x =0for all x ===.> a=b=0) is

    (there exist a,b) [~(ae^x + be^2x = 0 for all x ====> a=b=0)] ======>

    (there exist a,b) [ ae^x = be^2x =0 there exist x AND a=/=0 ,b=/=0]


    Whether you put the quantifier infront ,like i did, or at the end ,like you did we still have the same result.

    The negation of [tex]\forall xPx[/tex] is : [tex]\exists x\neg Px[/tex] .

    In our case Px is ( ae^x + be^2x =0 ======> a=b=0) and the negation of Px is:


    ae^x = be^2x =0 and a=/=0,b=/=0

    When you want to show that the function ae^x +be^2 is zero for all xεR ,YOU write :

    for all xεR, ae^x + be^2x =0 OR ae^x + be^2x = 0, for all x.

    To say that: for all x, is not part of the definition ,then what is part of??
     
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